Question #6fb36

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Question #6fb36

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Answer 1 · 2017-11-06T21:18:50

$1.99 \times 10^{- 18} J$ $1.99 xx 10^-18 " J"$

Explanation:

We know that $λ ν = c$ $lambda nu = c$ and $E = h ν$ $E = h nu$ , where

$λ$ $lambda$ is wavelength $(m)$ $("m")$

$ν$ $nu$ is frequency $(s^{- 1}$ $("s"^-1$ or $Hz)$ $"Hz")$

$c$ $c$ is the speed of light, $2.998 \times 10^{8} m/s$ $2.998 xx 10^8 " m/s"$

$h$ $h$ is Planck's constant, $6.626 \times 10^{- 34} J \cdot s$ $6.626 xx 10^-34 " J" * "s"$

We could directly use $E = h ν$ $E = h nu$ , but since we don't know $ν$ $nu$ , we can solve for it in the first equation and substitute.

$λ ν = c$ $lambda nu = c$

$ν = \frac{c}{λ}$ $nu = c/lambda$

So, $E = \frac{h c}{λ}$ $E = (hc) / lambda$ .

The given wavelength is $100. nm$ $"100. nm"$ . Convert this to meters.

$lambda = 100. cancel("nm") xx ("1 m") / ("10"^9 cancel("nm") )= 10.0 ^-7 " m"$

$E_("photon") = (hc) / lambda$

$E_("photon") = (6.626 xx 10^-34 " J" * "s" * 2.998 xx 10^8 " m/s") / (10.0 ^-7 " m")$

$E_("photon") = 1.99 xx 10^-18 " J"$

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Question #6fb36

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